Problem: A curve is defined by the parametric equations $x=4t^{-2}+4$ and $y=8t^{-1}$. What is $\dfrac{d^2y}{dx^2}$ in terms of $t$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{t^3}{8}$ (Choice B) B $t$ (Choice C) C $-\dfrac{t^4}{8}$ (Choice D) D $\dfrac{8t}{4t^4+4}$
Solution: We are asked to find the second derivative of a parametric function. Recall that the first derivative of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ is found with the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ Then, the second derivative is found with this following rule: $\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{\dfrac{d}{dt}\left(\dfrac{v'(t)}{u'(t)}\right)}{u'(t)}$ Let's start by finding $\dfrac{dy}{dx}$. $\dfrac{dy}{dx}=t$ Now we can find $\dfrac{d^2y}{dx^2}$. $\begin{aligned} \dfrac{d^2y}{dx^2}&=\dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\left(\dfrac{dx}{dt}\right)} \\\\ &=\dfrac{\dfrac{d}{dt}\left(t\right)}{\dfrac{d}{dt}(4t^{-2}+4)} \\\\ &=\dfrac{1}{-8t^{-3}} \\\\ &=-\dfrac{t^3}{8} \end{aligned}$ In conclusion, $\dfrac{d^2y}{dx^2}=-\dfrac{t^3}{8}$.